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8.1.IDENTIFY
and
SET UP:
.
p
mv
=
2
1
2
.
K
mv
=
EXECUTE:
(a)
5
(10,000 kg)(12.0 m/s)
1.20 10 kg m/s
p
==
×
⋅
(b)
(i)
5
60.0 m/s
2000 kg
p
v
m
×⋅
=
. (ii)
22
11
TT
S
U
VS
U
V
mv
m v
=
, so
T
SUV
T
SUV
10,000 kg
(12.0 m/s)
26.8 m/s
2000 kg
m
vv
m
=
E
VALUATE
:
The
SUV must have less speed to have the same kinetic energy as the
truck than to have the same momentum as the truck.
8.7.IDENTIFY:
The
average force on an object and the object’s change in momentum are related by Eq. 8.9.
The weight of the ball is
.
wm
g
=
SET UP:
Let +
x
be in the direction of the final velocity of the ball, so
1
0
x
v
=
and
.
2
25.0 m/s
x
v
=
EXECUTE:
av
2
1
2
1
()
(
)
x
x
Ft
tm
vm
v
−=
−
x
gives
21
av
3
(0.0450 kg)(25.0 m/s)
5
6
2
N
2.00 10 s
xx
x
mv
mv
F
tt
−
−
=
−×
.
. The force exerted by the club is much greater than the weight
of the ball, so the effect of the weight of the ball during the time of contact is not significant.
2
(0.0450 kg)(9.80 m/s )
0.441 N
w
E
VALUATE
:
Forces
exerted during collisions typically are very large but act for a short
time.
8.9.IDENTIFY:
Use
Eq. 8.9. We know the intial momentum and the impluse so can solve for the final
momentum and then the final velocity.
SET UP:
Take the
x
axis to be toward the right, so
Use Eq. 8.5 to calculate the
impulse, since the force is constant.
1
3.00 m/s.
x
v
=+
EXECUTE:
(a)
x
J
pp
=−
(
)
( 25.0 N)(0.050 s)
1.25 kg m/s
JF
=
+
=
+
⋅
Thus
1.25 kg m/s
(0.160 kg)( 3.00 m/s)
x
pJ
p
=+ =
+
⋅ +
+
=
1.73 kg m/s
+
⋅
2
2
1.73 kg m/s
10.8 kg m/s (to the right)
0.160
kg
x
x
p
v
m
⋅
=
+
⋅
(b)
(negative since force is to left)
(
)
( 12.0 N)(0.050 s)
0.600 kg m/s
=
−
=
−
⋅
0.600 kg m/s
(0.160 kg)( 3.00 m/s)
0.120 kg m/s
x
p
−
+
=
−
⋅
2
2
0.120 kg m/s
0.75 m/s (to the left)
0.160 kg
x
x
p
v
m
−⋅
=
−
E
VALUATE
:
In part (a) the impulse and initial momentum are in the same direction and
x
v
increases. In part (b) the impulse and initial momentum are in opposite directions and
the velocity decreases.
8.16.IDENTIFY:
Apply
conservation of momentum to the system of you and the ball. In part (a) both objects
have the same final velocity.
SET UP:
Let +
x
be in the direction the ball is traveling initially.
(ball).
(you).
0.400 kg
A
m
=
70.0 kg
B
m
=
EXECUTE:
(a)
12
x
x
PP
=
gives
and
.
2
(0.400 kg)(10.0 m/s)
(0.400 kg
70.0 kg)
v
2
0.0568 m/s
v
=
(b)
x
x
=
gives
2
(0.400 kg)(10.0 m/s)
(0.400 kg)( 8.00 m/s)
(70.0 kg)
B
v
+
and
.
2
0.103 m/s
B
v
=
EVALUATE:
When the ball bounces off
it has a greater change in momentum and you acquire a greater
final speed.
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 Fall '08
 Kiel

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